Suppose we have a game where, at each turn, a person takes either one step forwa
ID: 3340539 • Letter: S
Question
Suppose we have a game where, at each turn, a person takes either one step forward or one step back. Xi measures how many steps the person is ahead of the origin (that is, negative values indicate the person is behind the origin) at the ith step. At the first trial X0 = 0. The probability of moving forward (that is, Xi = Xi1 + 1), pf are as follows.pf=
0.25 if Xi>0 0.5 if Xi=o
0.6 if Xi < 0
(a) Give the probability distribution for X2. (b) Give the probability distribution for X3. (c) What are the expected value and variance of X3? (d) Give the probability distribution for X3 if X1 = 1. (e) What are the expected value and variance of X3 if X1 = 1? Suppose we have a game where, at each turn, a person takes either one step forward or one step back. Xi measures how many steps the person is ahead of the origin (that is, negative values indicate the person is behind the origin) at the ith step. At the first trial X0 = 0. The probability of moving forward (that is, Xi = Xi1 + 1), pf are as follows.
pf=
0.25 if Xi>0 0.5 if Xi=o
0.6 if Xi < 0
(a) Give the probability distribution for X2. (b) Give the probability distribution for X3. (c) What are the expected value and variance of X3? (d) Give the probability distribution for X3 if X1 = 1. (e) What are the expected value and variance of X3 if X1 = 1? Suppose we have a game where, at each turn, a person takes either one step forward or one step back. Xi measures how many steps the person is ahead of the origin (that is, negative values indicate the person is behind the origin) at the ith step. At the first trial X0 = 0. The probability of moving forward (that is, Xi = Xi1 + 1), pf are as follows.
pf=
0.25 if Xi>0 0.5 if Xi=o
0.6 if Xi < 0
(a) Give the probability distribution for X2. (b) Give the probability distribution for X3. (c) What are the expected value and variance of X3? (d) Give the probability distribution for X3 if X1 = 1. (e) What are the expected value and variance of X3 if X1 = 1?
Explanation / Answer
(a)
X2 measures steps the person is from origin in 2 steps.
Case 1: Go forward in each 2 step
X2 = 2 with p = 0.5 * 0.25 = 0.125
Case 2: Go forward in 1 step and backward in another step (and viceversa)
X2 = 0 with p = 0.5 * 0.75 + 0.5 * 0.6 = 0.675
Case 3: Go backward in each 2 step
X2 = -2 with p = 0.5 * 0.4 = 0.2
So, PMF for X2 is,
X2 = 2 with p = 0.125
X2 = 0 with p = 0.675
X2 = -2 with p = 0.2
(b)
X2 measures steps the person is from origin in 3 steps.
Case 1: Go backward in each 3 steps
X3 = -3 with p = 0.5 * 0.4 * 0.4 = 0.08
Case 2: Go backward 2 steps and then 1 forward
X3 = -1 with p = 0.5 * 0.4 * 0.6 = 0.12
Case 3: Go backward 1 step and then 1 forward and then 1 backward
X3 = -1 with p = 0.5 * 0.6 * 0.5 = 0.15
Case 4: Go backward 1 step and then 2 forward
X3 = 1 with p = 0.5 * 0.6 * 0.5 = 0.15
Case 5: Go forward 1 step and then 2 backward
X3 = -1 with p = 0.5 * 0.75 * 0.5 = 0.1875
Case 6: Go forward 1 step and then 1 backward and then 1 forward
X3 = 1 with p = 0.5 * 0.75 * 0.5 = 0.1875
Case 7: Go forward 2 step and then 1 backward
X3 = 1 with p = 0.5 * 0.25 * 0.75 = 0.09375
Case 8: Go forward in each 3 step
X2 = 3 with p = 0.5 * 0.25 * 0.25 = 0.03125
So, PMF for X3 is,
X3 = 3 with p = 0.03125
X3 = 1 with p = 0.09375+0.1875+0.15 = 0.43125
X3 = -1 with p = 0.1875+0.15+0.12 = 0.4575
X3 = -3 with p = 0.08
(c)
Expected value is, E[X3] = 3 * 0.03125 + 1* 0.43125 - 1 * 0.4575 - 3 * 0.08 = -0.1725
E[X32] = 32 * 0.03125 + 12* 0.43125 + (-1)2 * 0.4575 + (-3)2 * 0.08 = 1.89
Var[X3] = E[X32] - (E[X3])2 = 1.89 - (-0.1725)2 = 1.860244
(d)
X1 = 1
Case 1: Go forward in each 2 step
X3 = 3 with p = 0.25 * 0.25 = 0.0625
Case 2: Go forward in 1 step and backward in another step (and viceversa)
X3 = 1 with p = 0.25 * 0.75 + 0.75 * 0.5 = 0.5625
Case 3: Go backward in each 2 step
X3 = -1 with p = 0.75 * 0.5 = 0.375
So, PMF for X3 is,
X3 = 3 with p = 0.0625
X3 = 1 with p = 0.5625
X3 = -1 with p = 0.375
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